We know that if an operator has $L^2$-kernel, then it is Hilbert-Schmidt. Is there a similar simple criterion to detect compact operators?

In particular, I'd like to know the following: Let $f$ be a Schwartz function on ${\mathbb R}^2$ with $\mathrm{supp}(f)\subset{\mathbb R}\times J$ for some compact Interval $J$. Let $$ k(x,y)=f(e^x,x-y) $$ Is the operator $T:L^2({\mathbb R})\to L^2({\mathbb R})$, $T(\phi)(x)=\int_{\mathbb R}k(x,y)\phi(y)\,dy$ a compact operator?

  • $\begingroup$ If you take $f(x,y)=A(x)B(y)$ then $T\phi (x)=A(e^x) (B*\phi)(x)$ could be not in $L^2$ unless you assume more conditions on $A$. $\endgroup$ Jul 22 at 10:16
  • $\begingroup$ @GiorgioMetafune: I don't see, why this function should not be in $L^2$. The convolution is in $L^2$ and you simply multiply with a bounded function, so it stays in $L^2$. $\endgroup$
    – Zero
    Jul 22 at 12:49
  • $\begingroup$ Sorry, you are right, I forgot the hypotheses on $f$. How do you see that $T$ is bounded in $L^2$? $\endgroup$ Jul 22 at 12:57
  • $\begingroup$ @GiorgioMetafune: If we split $f$ into $f^+$ and $f^-$, then one can see that $T$ is the difference of two positive operators on $L^2$; positive operators are always bounded. $\endgroup$ Jul 22 at 17:21
  • $\begingroup$ @Jochen Glueck Thanks, but at that time I was even doubting that $T$ acts on $L^2$. $\endgroup$ Jul 22 at 17:33

The answer is no, in general. Assume for example that $J=[0,1]$ so that $|f| \leq C \chi_{\bf R \times [0,1]}$. Then $|T\phi(x)| \leq C \int_{x-1}^x |\phi(y)|\, dy \leq C\int_0^1 |\phi(x+y)|\, dy$ and Minkowsky inequaility for integrals gives $\|T\phi_2\| \leq C\|\phi\|_2$. So boundedness follows without any smoothness assumption. Assume now that $f=\chi_{[0,1]\times[0,1]}$. Then $T\phi(x)=\int_{x-1}^x \phi (y)\, dy$ for $x \leq e$ and this is not a compact operator (consider $\chi_{[-n-1,-n]}$). To get a smooth counterexample it suffices to consider a smooth f with compact support which dominates $\chi_{[0,1]\times[0,1]}$. If $S$ is the corresponding integral operator, then $S\phi \geq T\phi$ for positive $\phi$ and $S$ is not compact.


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